Thread: destructor not called

Sorry, but I cannot reproduce the behaviour you described. The program prints five "constructing (...)" messages, I enter the integer input, and then it prints five "destructing (...)" messages, as expected.

Originally Posted by rahulsk1947

any way to see the destructor being called?

Run your program from the command line (e.g., open a command prompt window).

Alternatively, you can introduce a block for local scope:

Code:

int main(void)
{
    {
        Base b1(1),b2(2),b3(3);
        static Base b4(4);
        Base b5(5);
    }

    int exit;
    cin>>exit;
}

which would also help you to more clearly see the difference between the lifetime of a local and static object.

Originally Posted by legit

Sure, use std::cin.get() to force the program to wait for user input. std::cin.ignore() also works

That would have the same effect as the current code: the destructor messages will not be seen because they are presumably being shown on an IDE initiated command prompt that disappears very quickly after the program terminates, but if the program has not terminated, they will not be printed because the objects have not been destroyed.

Also, if this is a base class it would be a good idea to make the destructor virtual.

Example:

Code:

struct unsafe_base
{
	~unsafe_base( void )
	{
		cout << "~unsafe_base( )" << endl;
	}
};

struct unsafe_derived : unsafe_base
{	
	~unsafe_derived( void )
	{
		cout << "~unsafe_derived( )" << endl;
	}
};

struct safe_base
{
	virtual ~safe_base( void )
	{
		cout << "~safe_base( )" << endl;
	}
};

struct safe_derived : safe_base
{	
	~safe_derived( void )
	{
		cout << "~safe_derived( )" << endl;
	}
};

int main( void )
{
	auto_ptr< unsafe_base >
		u( new unsafe_derived( ) );
	auto_ptr< safe_base >
		s( new safe_derived( ) );
	return 0;	
}